How many three-digit numbers are divisible by 7?

Solution:

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore, 105, 112, 119,\ldots

All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.

The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.
Clearly, 999 - 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.

$105,112,119, \ldots \ldots, 994$

Let 994 be the $\mathrm{n}^{\text {th }}$ term of this A.P.

a = 105

d = 7

$a_{n}=994$

$\mathrm{n}=?$

$a_{n}=a+(n-1) d$

994 = 105 + (n – 1) 7

889 = (n – 1) 7

(n – 1) = 127

n = 128

Therefore, 128 three-digit numbers are divisible by 7.